Quadratic Equations

Quadratic equation

An equation of the form ax² + bx + c = 0 with a, b and c real and a ≠ 0, called its standard form. More generally, setting any degree-2 polynomial p(x) equal to zero gives a quadratic equation.

Root of a quadratic equation

A real number α for which aα² + bα + c = 0; α is then also called a solution of the equation. The roots of ax² + bx + c = 0 are exactly the zeroes of the polynomial ax² + bx + c.

Why can a quadratic equation never have more than two roots?

Its roots are the zeroes of a quadratic polynomial, and a polynomial of degree 2 has at most two zeroes. So no quadratic equation can be satisfied by three different real numbers.

How do you decide whether an equation such as x(x + 1) + 8 = (x + 2)(x − 2) is quadratic?

Expand and simplify both sides before judging. Here the x² terms cancel, leaving x + 12 = 0, so it is not quadratic. In the chapter's other example, (x + 2)³ = x³ − 4 looks cubic, but the x³ terms cancel and it reduces to x² + 2x + 2 = 0, which is quadratic.

How do you solve a quadratic equation by factorisation?

Split the middle term so the quadratic becomes a product of two linear factors, then set each factor equal to zero. In the chapter's example, 2x² − 5x + 3 = 2x² − 2x − 3x + 3 = (2x − 3)(x − 1), so the roots are 3/2 and 1.

When factorising 6x² − x − 2, how do you choose how to split the middle term?

Look for two terms whose sum is the middle term and whose product equals (x² coefficient × constant) x², here −12x². So −x becomes 3x − 4x, giving 3x(2x + 1) − 2(2x + 1) = (3x − 2)(2x + 1) and roots 2/3 and −1/2.

What does it mean when a quadratic factorises into the same linear factor twice?

The equation has a single value repeated as both roots, one for each copy of the factor. In the chapter's example, 3x² − 2√6x + 2 = (√3x − √2)(√3x − √2), so both roots are √2/√3.

How do you turn the prayer hall problem, carpet area 300 m² and length one more than twice the breadth, into a quadratic and solve it?

Let the breadth be x m, so the length is (2x + 1) m and x(2x + 1) = 300, i.e. 2x² + x − 300 = 0. Factorising gives (x − 12)(2x + 25) = 0. A breadth cannot be negative, so it is 12 m and the length 25 m.

A train covers 480 km at a uniform speed; 8 km/h slower, it would take 3 hours more. How do you set up the quadratic equation?

Let the usual speed be x km/h. Time is distance ÷ speed, so the times are 480/x and 480/(x − 8) hours, and 480/(x − 8) − 480/x = 3. Clear the denominators to get a quadratic in x, then reject any root that is not a possible speed.

Why do you sometimes discard one root of a quadratic equation in a word problem?

Both roots satisfy the equation, but the unknown may stand for something that must be positive, such as a breadth or a distance. A negative root, like −12.5 m for the prayer hall's breadth or −12 m for the pole's distance, has no meaning there.

Quadratic formula

The roots of ax² + bx + c = 0 are x = (−b ± √(b² − 4ac)) / 2a, provided b² − 4ac ≥ 0.

Discriminant

The quantity b² − 4ac for the quadratic equation ax² + bx + c = 0. It gets this name because its sign determines whether the equation has real roots and whether they are distinct or equal.

Condition for two distinct real roots

b² − 4ac > 0. The roots are then (−b + √(b² − 4ac))/2a and (−b − √(b² − 4ac))/2a, which differ because the square root is non-zero.

Condition for two equal real roots

b² − 4ac = 0. The square root vanishes, so both roots equal −b/2a; these are also called coincident roots.

Condition for no real roots

b² − 4ac < 0. No real number has a negative square, so √(b² − 4ac) is not real and the equation has no real roots.

How do you find the nature of the roots of 2x² − 4x + 3 = 0 without solving it?

Identify a = 2, b = −4 and c = 3, then compute the discriminant: (−4)² − 4 × 2 × 3 = 16 − 24 = −8. It is negative, so the equation has no real roots.

How does the discriminant decide whether a situation like the chapter's pole in a circular park is possible?

Model it as a quadratic and check the discriminant before solving. The pole gives x² + 7x − 60 = 0 with discriminant 289 > 0, so real positions exist; the formula gives x = 5, placing the pole 5 m from one gate and 12 m from the other.

How do you find the value of k for which a quadratic such as 2x² + kx + 3 = 0 has two equal roots?

Write the discriminant in terms of k, set it equal to zero because equal roots need b² − 4ac = 0, and solve that equation for k. Any value of k it gives makes the roots coincide.

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